Round Numbers((组合数 + 简单组合数学)||(数位dp))

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概述

Round Numbers

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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Description

The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (also known as 'Rock, Paper, Scissors', 'Ro, Sham, Bo', and a host of other names) in order to make arbitrary decisions such as who gets to be milked first. They can't even flip a coin because it's so hard to toss using hooves.

They have thus resorted to "round number" matching. The first cow picks an integer less than two billion. The second cow does the same. If the numbers are both "round numbers", the first cow wins,
otherwise the second cow wins.

A positive integer N is said to be a "round number" if the binary representation of N has as many or more zeroes than it has ones. For example, the integer 9, when written in binary form, is 1001. 1001 has two zeroes and two ones; thus, 9 is a round number. The integer 26 is 11010 in binary; since it has two zeroes and three ones, it is not a round number.

Obviously, it takes cows a while to convert numbers to binary, so the winner takes a while to determine. Bessie wants to cheat and thinks she can do that if she knows how many "round numbers" are in a given range.

Help her by writing a program that tells how many round numbers appear in the inclusive range given by the input (1 ≤ Start < Finish ≤ 2,000,000,000).

Input

Line 1: Two space-separated integers, respectively Start and Finish.

Output

Line 1: A single integer that is the count of round numbers in the inclusive range Start.. Finish

Sample Input

2 12

Sample Output

题目意思：

给定一个2 * 10^9范围内的一个范围L,R 让你在这个范围内有多少个十进制数字中的二进制0的个数大于等于1的个数

题解：

可以设定一个get_prefix(int n)的函数，也就是求出从1---->n中有多少个二进制0的个数大于等于1的个数的合

法数字，然后减一下就可以。

具体求的方法是，把n转换为2进制，假设为n的二进制一共有len这么长，先不考虑n的二进制最高位，即只考虑剩下的

len-1位，当长度为len-1时，最高位为1，剩下的便是一个计数组合问题了，放多少个0是合法的，即1XXXXX....

具体的例子借鉴一下Bin神题解的例子是22

拆为二进制为 22 = 1 0 1 1 0  len = 5

去除最高位的1，

当len = 4的时候，最高位为1，即1XXX，有三个位置可以放，可以放0也可以放1，合法的放法是2个0或3个0(算上最

高位的1)，即有C(3,2) + C(3,3) = 4

同理，len = 3的时候，最高位为1，即1XX，有两个位置可以放，有C(2,2) = 1

同理，len = 2的时候，有C(1,1) = 1

剩下要解决的便是len = 5的时候的，要根据n的二进制去遍历n的二进制数组，当较高位为0的时候可以略过，担当最高

位为1的时候就可以把1换为0了.

具体见代码

#include <algorithm>
#include <iostream>
#include <numeric>
#include <cstring>
#include <iomanip>
#include <string>
#include <vector>
#include <cstdio>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#define LL long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const LL MAX = 405;
const double esp = 1e-6;
const double PI = 3.1415926535898;
const int INF = 0x3f3f3f3f;
using namespace std;
LL dp[40][40];
void init(){
    memset(dp,0,sizeof(dp));
    for(int i=0;i<=35;i++)
        dp[i][i] = dp[0][i] = dp[1][i] = 1;
    for(int i=0;i<=35;i++)
        dp[i][0] = 1;
    for(int i=1;i<=35;i++){
        for(int j=1;j<=35;j++){
            if(i != j)
                dp[i][j] = dp[i-1][j] + dp[i-1][j-1];
        }
    }
}
LL cal(LL n){
    if(n <= 1)
        return 0;
    LL two[40],t = 0,m = n,zero = 0,one = 0;
    while(m){
        two[t++] = m%2;
        if(m%2)
            one += 1;
        else
            zero += 1;
        m /= 2;
    }
    LL ans = 0;
    for(int i = t - 1;i > 0;i--){
        if(i%2)
            ans += ((1<<(i-1)) - dp[i-1][(i-1)/2]) / 2;
        else
            ans += ((1<<(i-1))) / 2;
    }
    if(zero >= one)
        ans += 1;
    zero = 0;one = 1;
    for(int i =  t - 2;i >= 0;i--){
        if(two[i] == 0)
            zero += 1;
        else{
            for(int j=i;j >= 0 && j + zero + 1 >= i - j + one;j--)
                ans += dp[i][j];
            one += 1;
        }
    }
    return ans;
}
int main(){
    LL n,m;
    init();
    while(~scanf("%lld %lld",&n,&m)){
        printf("%lldn",cal(m) - cal(n - 1));
    }
    return 0;
}